Question

An unknown voltage source $E_u$ (negligible internal resistance) is connected to a potentiometer circuit as shown in the following figure. If the galvanometer current is $10\mu A$ with the direction as indicated, then value of $E_u$ is

• A. 0.9910 V
• B. 0.9991 V
• C. 1.0012 V
• D. 1.0048 V

Answer 1

The correct option is D 1.0048 V


$I_g=10\mu A={10}^{-5}A$

writing KVL in Loop 1, we get

$1.6-100I-500I-1000(I+I_g)=0$

$1.6-1000I_g=1600I$

$1.6-1000\times {10}^{-5}=1600I$

$I=\frac{1.6-0.01}{1600}=9.9375\times {10}^{-4}A$

Now in loop 2,

$E_u={10}^{-5}\times 100+1000\times {10}^{-5}+1000\times 9.9375\times {10}^{-4}$

$=1100\times {10}^{-5}+0.99375$

= 0.011 + 0.99375

= 1.00475 = 1.0048 V